I've been thinking about the following problem:
Let $B$ be matrix whose off-diagonal entries are non-negative (possibly zero) and whose rows sum to zero, where there is some $t$ such that $\exp(tB)$ is entry positive, then how can we argue that $\exp(tB)$ is entrywise positive for each $t > 0$?
One way for me to assert the claim would be to show that $B$ off diagonal entries are positive, which would imply the result by this result: A matrix whose off-diagonal entries are $>0$ has its exponential with all positive entries
I'm not quite sure what information exactly, $\exp(tB)_{ij} >0$ gives me.
Best Answer
Let $m = \min_i(B_{ii}) - 1$. Then $tB - tmI$ is entry-wise positive for $t > 0$. By the Taylor expansion definition of the matrix exponential, $e^{tB - tmI}$ is entry-wise positive as well. Since $I$ and $B$ commute, $e^{tB} = e^{(tB-tmI) + tmI} = e^{tB - tmI}e^{tm}$. Hence $e^{tB}$ is also entry-wise positive.
(Note that you say $t=0$ is allowed, though it should not be. $e^0 = I$ has zeros on the off-diagonals.)
Edit: I see that $B$ was actually assumed to have non-negative off-diagonal entries, not positive entries, so there's a little more work to do.
We're given that there is $t_0 > 0$ such that $e^{t_0 B}$ has positive entries. That means $e^{t_0(B - mI)}$ has positive entries. Examining the Taylor series, that means for every $i,j$, there exists $k$ such that the $i,j$ entry of $(B-mI)^k$ is positive. But that in turn means $e^{t(B - mI)}$ has positive $i,j$ entry for any $t > 0$, so $e^{tB}$ has positive entries.